General > General Technical Chat
Powering an LED from international line voltage
soldar:
How about the classic NE-2 indicator lamp?
https://en.wikipedia.org/wiki/Neon_lamp
Ian.M:
Its certainly not impossible to stabilise the LED current from a capacitive dropper, but the problem is: the capacitor, line voltage and frequency pretty much set the current, and if you want less you either have to burn it off as heat in a shunt regulator, or use a series regulator that has to stand-off a significant fraction of the peak mains voltage. You probably wont like the resulting cost and complexity (especially vs a NE2 neon + dropper resistor)!
See attached LTspice sim for a slightly different take on the shunt regulator option using jellybean parts apart from the dropper capacitor. It stabilises the LED current at about 15mA with under 2mA variation between 100V in and 260V in. As its full-wave and tries to reduce flicker, there are considerable options to Muntz it if you don't care about flicker.
mikeselectricstuff:
Capacitive dropper plus some resistance to make the current more consistent- this will not be constant current but may be good enough - does brightness change over supply voltage really matter ?
Also, do you really need 10-15mA ? A more efficient LED/Color (e.g. white or green) may make the problem easier by reducing current, also consider multiple LEDs in series to reduce the current.
Gyro:
--- Quote from: jonpaul on May 03, 2024, 06:12:32 am ---Use any VDE/TUV /CE approved led driver, cost ~ EU 5..15 depending on the LED power, V, I.
We use the Italian TCI
Avoid the Chinese drivers
j
--- End quote ---
EUR 5..15 to drive a 10-15mA indicator LED? Seriously? :o
@OP: I would suggest something like a mains dropper, feeding into a small package, low current bridge rectifier [Edit: or discrete diodes] (flicker removal), then a Zener of, say, 4V7 followed by a small resistor to limit the LED current down from 4V7.
Assuming an efficient LED @ 10mA, a resistive mains dropper would dissipate 750mW at 80V and 2.5W at 250V - much too high in my opinion. A capacitive dropper is the way to go. Assuming a 470nF safety rated capacitor, this would give ~12mA at 80V (Zener essentially dissipating nothing) and 37mA to 250V. At maximum voltage, the dissipation of the Zener, after taking off the 10mA for the LED would be only 127mW, so a 250mW rating would be fine. Don't forget to put a low value series resistor (preferably fusible) of around 100R as the first component between mains live and the capacitor to cope with mains spikes.
EDIT: My capacitor reactance calculation was for 50Hz. At 60Hz, you would get away with a 390nF cap, maybe even 330nF.
nctnico:
--- Quote from: mikeselectricstuff on May 03, 2024, 09:10:09 am ---Capacitive dropper plus some resistance to make the current more consistent- this will not be constant current but may be good enough - does brightness change over supply voltage really matter ?
Also, do you really need 10-15mA ? A more efficient LED/Color (e.g. white or green) may make the problem easier by reducing current, also consider multiple LEDs in series to reduce the current.
--- End quote ---
I agree. A capacitive dropper is the easiest way. An anti-parallel diode across the LED (or a dual LED in one package) is necessary to make an AC current flow. Not much current is needed for a modern LED so designing something which works from 80VAC to 260VAC should be possible.
Navigation
[0] Message Index
[#] Next page
[*] Previous page
Go to full version